Let $m = (9n^2 + 54n + 80)(9n^2 + 45n + 54)(9n^2 + 36n + 35)$. The greatest positive integer which divides $m$ for all positive integers $n$ is:

Team $A$ consists of $7$ boys and $n$ girls and Team $B$ has $4$ boys and $6$ girls. If a total of $52$ single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl,then $n$ is equal to

All possible $5$-digit numbers,each having $5$ distinct digits,are formed using the digits $\{1, 2, 3, 5, 6, 8\}$. Among them,the number of such numbers which are divisible by $3$ but not by $6$ is:

Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example,$2210222$ is in $S$,but $0210222$ is not in $S$. Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$,is equal to $....$

The number of non-negative integer solutions of the equations $6x + 4y + z = 200$ and $x + y + z = 100$ is

The value of $\sum_{r=0}^{20} {}^{50-r}C_{6}$ is equal to